KDB中lateral join的等价操作及数组值列展开方法咨询

Hey there! Let's tackle your questions about kdb+ equivalents to PostgreSQL's lateral join and unnesting array columns—they're closely linked, so I'll walk you through both with concrete examples.

1. The Direct Lateral Join Equivalent: ungroup #

In PostgreSQL, you use lateral join (often paired with unnest) to split a row with an array column into multiple rows—one for each element in the array—while preserving all other original row values.

In kdb+, the ungroup function does exactly this out of the box. Let's use your table T with array column C as an example:

First, create a sample table to test with:

T: ([] id: 1 2 3; C: (10 20; 30; 40 50 60))

To unnest column C (matching the behavior of a PostgreSQL lateral join with unnest), simply run:

ungroup T

This outputs:

id C
-----
1  10
1  20
2  30
3  40
3  50
3  60

This is exactly the result you'd get from a PostgreSQL lateral join unnesting the array column.

If your table has multiple array columns and you only want to unnest one, specify it explicitly:

ungroup[;`C] T

2. Flexible Custom Logic: Using each #

If you need to add custom processing while unnesting (like you might do with a subquery in a lateral join), use the each iterator to handle each row individually, then combine the results.

For example, let's unnest C and calculate a new value using each array element and the row's id:

// Process each row to create expanded rows with custom logic
expandedRows: {[row] 
    ([] id: enlist row[`id]; C: row[`C]; calculated: row[`C] + row[`id]) 
} each T

// Combine all expanded rows into a single table
,/ expandedRows

This gives you:

id C  calculated
----------------
1  10 11
1  20 21
2  30 32
3  40 43
3  50 53
3  60 63

This mimics the flexibility of a lateral join where you can run custom subqueries per row.

Quick Tips #

• If your array column contains empty lists, ungroup will drop those rows entirely. To keep them (with a null value for the array element), first replace empty lists with a single null element:

  T: update C: $[0=count each C; enlist ::; C] from T
  ungroup T
  

内容的提问来源于stack exchange，提问作者JSLover